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I was looking at this Wikipedia article about the characteristic equation of a linear differential equation.

I understand why a linear combination of any solutions that are already found must also be a solution, but I don't know why we can be sure that this yields all possible solutions. (In other words, why do we know that a linear ODE with degree n must have n linearly independent solutions, and is this true for all ODEs?)

Agastya Ravuri
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  • There are existence and uniqueness theorems. For linear ODEs of degree $n$, the solution depends on the $n$ parameters $y(0), y'(0), ..., y^{(n-1)}(0)$. You can turn such problems into vector ODEs $\vec{x}'(t) = A\vec{x}(t)$ with $\vec{x}(t) = (y(t), y'(t), ..., y^{(n-1)}(t))$ where each $\vec{x}(0) \in \mathbb{R}^n$ yields a unique solution. If $\vec{x}1(t), ..., \vec{x}_n(t)$ are $n$ solutions with linearly independent ${\vec{x}_1(0),...,\vec{x}_n(0)} $ then $$ \left(\sum{i=1}^na_i\vec{x}i(t)=0\quad \forall t \right)\implies \sum{i=1}^na_i\vec{x}_i(0)=0 \implies a_i=0 \quad \forall i$$ – Michael Jan 02 '23 at 22:55

  • You can turn such problems into vector ODEs

    How do you do this? Is the matrix A something like the matrix ((0, 1, 0, ... , 0, 0), (0, 0, 1, ... , 0, 0), ... (0, 0, 0, ..., 1 ,0), ([coefficients from the equation]))? If so, why does this mean each x(0) must yield a unique solution (and this must be all solutions)? Thanks for the help.

     – Agastya Ravuri Jan 02 '23 at 23:21
  • The matrix $A$ is as you say. Uniqueness for each $x(0) \in \mathbb{R}^n$ can be shown by showing that the all-zero initial condition $0 \in \mathbb{R}^n$ cannot spring forth spontaneous life. Existence for each $x(0) \in \mathbb{R}^n$ is intuitive by fixing $\delta>0$ very small and using the iteration $x((k+1)\delta) \approx x(k\delta) + \delta Ax(k\delta)$ for $k \in {0, 1, 2, ...}$. The existence and uniqueness theorems formalize this. If $A$ has a linear independent collection of eigenvectors $v_i$ then you can explicitly get existence by $x(t) = \sum_{i=1}^n a_i v_i e^{\lambda_i t}$. – Michael Jan 03 '23 at 01:18
  • see this answer. – Abolfazl Chaman motlagh Jan 03 '23 at 20:07

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